Kendra has $5.45 in nickels and quarters. if she has 11 more quarters than nickels, how many of each coin does she have?

If Kendra were to add 11 more nickels, she would have equal numbers of quarters and nickels, so a multiple of $0.30. Adding $0.55 to her total would give her $5.45 + 0.55 = $6.00, which is 20 times $0.30. At that point she would have 20 quarters and 20 nickels. But we know 20 nickels is 11 more than the number she really has, so ...

Kendra has 9 nickels and 20 quarters.

_____

If you want to write equations, you can let n and q stand for the numbers of nickels and quarters Kendra has.

... 5n +25q = 545 . . . . . the amount of money Kendra has

... q = n+11 . . . . . . . . . . . the relationship between numbers of quarters and nickels

Using the second equation to substitute for q in the first, we have

... 5n +25(n+11) = 545

... 30n +275 = 545

... 30n = 270

... n = 270/30 = 9

... q = n + 11 = 9 + 11 = 20

If you were to work this problem substituting for n instead of q, you would find the equations match the logic used in the text description above.

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

dont think so

step-by-step explanation: